so welcome to the lecture series on mathematical methods and their applications uh in the last
lecture in the last few lectures we have seen that what laplace transforms are what are
their properties and how can we solve an ordinary differential equation or a partial differential
equation ontical equations using laplace transforms now the next topic is uh fourier series now
in this uh lectures we will see that what fourier series are and uh what is the convergence
theorem for fourier series and what are the applications of fourier series or fourier
integrals ok now uh first few introduction first introduction on uh fourier series it
is named after a french mathematician and a physicist jacques fourier who was the first
to use fourier series in his work a series expansion of a function in terms of a trigonometric
functions cos m x and sin n x is called a fourier series so fourier series is nothing
but if you express a function in terms of sine and cosines then we call such a series
as fourier series many functions including some discontinuous periodic functions can
be written in the fourier series and hence it has a wide application in solving ordinary
and partial differential equations now let us start with this set set is one
cos pi x by l cos two pi x by l and so on sine pi x by l sine two pi x by l and so on
now what is an orthogonal set of functions how do we define it suppose we have some functions
suppose functions are f one f two f three and so on and it is defined suppose defined
in a comma b then we say that this set of function is orthogonal if in this interval
if integral a to b f i x into f j x d x is equals to zero for all i not equal to j so
if this condition hold if this condition hold for set of functions f one f two up to uh
f three and so on which is defined in the interval a comma b then we say that the set
of functions are orthogonal ok now consider this set one cos pi x by l cos two pi x by
l and so on sin pi x by l sin two pi x by l and so on now in the interval minus l to
l if we see if we carefully see then minus integral minus l to l cos m pi x by l d x
and this equal to zero again the second inequality holds the second inequality also hold it is
zero when m not equal to n the next inequality minus l to l sin m pi x by l and sin n pi
x by l d x equal to zero when m not equal to n and is l when m equal to n so this we
can derive very easily and minus l to l cos pi m pi x by l into this is also zero for
all m and n so hence if we take any two multiples in this
set in this set if we take any two different functions and integrate from minus l to l
it is always zero so we say that this set of this set is an orthogonal set of functions
ok because it satisfies this condition it satisfies this condition because integral
a to b f i x f j x is zero for all i not equal to j and this set satisfies this property
hence we say that this set is nothing but an orthogonal set of functions now uh now
let f x be a periodic function of period two n and define an interval minus l to l ok assume
that now we are assuming that it is it can be expressed as the linear combination of
trigonometric functions of cos m x and sin m x so what i want to say so first i construct
a orthogonal set of functions and that set is nothing but this set one cos pi x by l
cos two pi x by l and so on sin pi x by l sin two pi x by l and so on ok this is an
orthogonal set of function and a function which is periodic ok in the interval minus
l having a period two l defined on minus l to l which is given can be expressed as a
linear combination of these functions that is some multiple of this we are taking for
convenience as a naught by two plus a one into this a two into this and so on b one
into this b two into this and so on so which can be written as summation n from one to
infinity a n cos n pi x by l plus summation n from one to infinity b n sin n pi x by l
ok this function we can write as a linear combination
of these trigonometric functions sine and cos now if you want to find out the values
of a naught a n b n and so on so how can you find these values so what is this expression
basically this is a naught by two plus a one cos pi x by l plus a two cos two pi x by l
and so on plus a n cos n pi x by l and so on this is what we are having here and here
it is b one sin pi x by l plus b two sin pi x sin two pi x by l sin two pi x by l and
so on plus b n sin n pi x by l and so on so basically the uh the linear combination of
these functions is this thing or this thing ok now to find out the value of a naught let
us integrate both the sides from minus l to plus l ok so if we integrate from minus l
to l in this side it is minus l to l f x d x is equals to a naught by two integral minus
l to plus l d x plus now integral minus l to l cos n pi x by l for every n is zero and
integral minus l to l sine n pi x by l d x is zero for any n so that means these all
are zero so this implies a naught is nothing but one by l times minus l to l f x d x
so that is how we can compute a naught one by l minus integral minus l to l f x d x ok
now suppose you want to compute a n a n uh n may be one two three and so on suppose you
want to compute this a n ok so you multiply both sides by cos n pi x by l ok so when you
multiply both sides by cos pi x by l it is f x into cos n pi x by l integrate from minus
l to plus l d x is equal to now here if you multiply and integrate from minus l to l cos
pi x by l is zero for any n now if it will be not zero cos into cos is not equal to zero
only when m equal to n that is n equal to n and all in all other cases it will be zero
and cos and sine multiple is always zero for any m and n this is always zero here what
we have here it is nothing but a n integral minus l to l cos square n pi x by l into d
x and we have already seen that this value is nothing but l a n into l because when m
equal to n so this value is nothing but l this we have already seen so from here we
can say that a n is nothing but one by l times integral minus l to l f x cos n pi x by l
into d x so this is how we can find out a n and what is n n may be one may be two may
be three and so on ok clear now uh now suppose you want to find out b
n in the same way you multiply both the sides by sin n pi x by l this is integral minus
l to l f x into sin n pi x by l into d x integrate both the sides here integral minus l to l
sin n pi x by l is zero and the multiple of sine and cos for any m and n integral minus
l to l is always zero so all these terms are zero and here we have and here when m is not
equal to n it is zero we left with only b n so integral minus l to l b n into sin square
n pi x by l into d x and this value is nothing but l when m equal to n so it is b n into
l so this implies b n is nothing but one by l integral minus l to l f x sin n pi x by
l into d x so these formulas are called basically eulers formula ok to find a naught a n and
b n so uh if we assume that if we express function periodic function f x as a linear
combination of cos trigonometric functions of cos and sine then a naught a n and b n
can be find using these expressions ok now let us solve some problems based on this
that how we can find fourier series of periodic function so basically uh so basically if we
have a periodic function f x of period two l defined in minus l to l so what will be
f x what we have learnt that f x is nothing but a naught by two plus summation n varying
from one to infinity a n cos n pi x by l plus summation n from one to infinity b n sin n
pi x by l now what are what is a naught a naught is nothing but one by l integral minus
l to l f x d x this we have derived a n is nothing but one by l integral minus l to l
f x cos n pi x by l into d x and what is b n it is one by l integral minus l to l f x
sin n pi x by l into d x here in these two expressions n is one two three and so on so
now let us solve this problem here it is given that function is periodic and has a period
two l oh sorry two pi ok and what is a what is function what is uh nature of function
function is pi plus x when x is varying from zero to zero to pi zero to minus pi and it
is zero when zero less than equal to x less than pi and f x plus two pi is f pi f x because
period is two pi so what is the shape of the function when it is minus pi so it is zero
and when it is zero it is pi so here it is like this and from zero to pi suppose pi is
here from zero to pi it is zero again from pi to two pi it is this function
say i am having the same height and from pi to three pi is again zero then three pi to
four pi again it is uh this line and four pi to five pi it is zero so this function
is something like this is the shape of this i mean graph of this function now how can
we find the fourier series of this function so so we will express this function in this
form a naught by two plus summation a n cos n pi x by l plus b n plus summation b n sin
n pi x by l now we first compute a naught and a n b n what is a naught a naught is nothing
but one by l now instead of l here we have pi because two l is two pi oh that means l
equal to pi it is one by pi integral minus pi to plus pi f x d x so this is nothing but
one by pi integral from minus pi to zero it is pi plus x otherwise it is zero [vocalized-noise]
so it is nothing but one by pi and this term is nothing but pi x plus x square upon two
minus pi to zero upper limit minus lower limit when you apply the lower limit it is nothing
but negative of pi into minus pi plus pi square upon two and it is minus pi square by two
that is pi by two so this is a naught ok so a naught for this
problem is pi by two so i am writing here a naught is nothing but pi by two for this
problem ok so a naught is pi by two now let us find a n a n a n is nothing but one by
pi integral minus pi to plus pi f x and it is cos n pi x by l l is pi into d x so pi
pi cancel out it is nothing but one by pi integral minus pi to zero pi plus x cos n
x d x which is nothing but one by pi now you will integrate this apply integration by parts
so it is first as it is integration of second sin n x upon n minus this derivative derivative
is one into integration of this that is minus cos n x upon n square from minus pi to zero
so this is nothing but one by pi when uh when x equal to zero sine is zero and when x equal
to pi or minus pi sin n pi is zero so this term is zero from the lower limit and the
upper limit so this term is zero when x is zero cos zero is one and when x is minus pi
it is cos n pi and cos n pi is uh cos n pi is minus one for negative n for odd n i mean
what is cos n pi cos n pi is nothing but minus one k to the power n when n is odd it is minus
one when n is even it is one so uh this is nothing but minus minus plus it is one by
n square minus cos n pi by n square so this is nothing but one by pi n square one minus
minus one k to the power n so this is this is a n for this problem so what is a n a n
will be nothing but one by n square pi one minus minus one k to the power n
now let us compute the last term that is b n b n will be nothing but again one by pi
integral minus pi to pi it is f x sin n pi x by l l is pi so this is one by pi again
integral minus pi to zero it is pi plus x by the definition of the function now again
you will integrate by parts here pi plus x sin n x ok integration of this so integration
of this is nothing but minus cos n x by n minus derivative of first integration of second
that is minus sin n x by n square from minus pi to zero now again this term when you take
x equal to zero it is zero when you take x equal to minus pi it is zero and here you
take you can take minus outside one by pi when you substitute x as minus pi it is zero
when you take x equal to zero it is uh it is pi by n ok when you take when you take
x equal to zero it is it is pi pi by n and when you take x equal to minus pi it is zero
so it is nothing but minus one by n ok so what will be so b n is nothing but minus one
by n so what will be the fourier series of this function
so fourier series representation of this function f x is nothing but f x can be written as a
naught by two what is a naught a naught is pi by two a naught by two that is pi by four
pi by four plus summation n from one to infinity a n a n is nothing but this expression so
it is one by n square pi one minus minus one k to the power n into cos n pi x by l plus
summation n from one to infinity it is b n sin n pi x by l what is b n minus one by n
it is minus one by n sin n pi x by l so which if you expand this it is nothing but pi by
four plus when n is minus when n is one it is nothing but zero when n is two ok when
n is one it is nothing but two for odd for even values of n it is zero so that means
from here only odd values come it is nothing but two will be outside two upon pi and it
is nothing but and this l is nothing but pi sorry this l is nothing but pi pi pi cancels
out because l is nothing but pi so you substitute when you substitute x as one so when x is
one it is two so two i am taking outside pi i am taking outside [FL] it is cos x upon
one square plus cos three square and so on and minus it is uh sin x upon one plus sin
two x upon two plus sin three x upon three so this will be the fourier series representation
of this function f x ok now let us solve one more problem based on
this uh fourier series representation of this function and hence reduce this series ok so
let us solve this problem also so here period is again two pi so l equal to pi ok and function
is x minus x square so directly compute a naught a n and b n after calculating a naught
a n and b n we substitute it over here find f x and hence the fourier series representation
of this function and then substituting some values of x we will try to obtain this uh
series ok so what is f x here f x is x minus x square now what will be a naught again one
by pi minus pi to plus pi f x d x it is one by pi minus pi to plus pi x minus x square
by two d x now this x is a odd function from minus pi to pi it is zero and this x square
is an even function so it will be two times so we can easily write it is minus two by
upon pi zero to pi x square d x ok and it is nothing but minus two upon pi it is x cube
upon three zero to pi so which is nothing but minus two upon pi pi cube by three and
it is equal to minus two by three pi square so this is a naught for this problem so i
am writing a naught over here so a naught is minus two by three pi square now let us
compute a n for this function so what will be a n a n is one by pi integral minus pi
to plus pi f x cos n pi x by l into d x ok now again x into this is odd function it will
be zero from minus pi to plus pi and x square into this is an even function so it will be
two times so we can write it here minus two upon pi integral zero to pi x square cos n
x d x so this is nothing but minus two upon pi you will integrate by parts x square sine
n x upon n minus two x this is minus cos n x upon n square then uh plus two this is minus
sin n x upon n cube and the whole expression from zero to pi now whatever terms of sine
from zero to pi is zero because upper limit is zero sin n pi is zero and sin zero is zero
this is zero this is zero now only this term lefts now it is also when x is zero is zero
so only the uh above limit left it is minus two upon pi it is two pi and it is cos n pi
upon n square which is nothing but pi pi cancel out so it is minus four by n square into minus
one k to the power n so a n is nothing but minus four upon n square minus one k to the
power n when n varying from one two three and so on ok now b n can be calculated in
the similar way so what will be b n b n will be one by pi integral minus pi to plus pi
x minus x square sin n pi x by l into d x now x into this is an even function and this
is an odd function this will be zero and we left with only two upon pi integral zero to
pi x sin n x d x now again integration by parts it is first
as it is integral second it is minus cos n x upon n minus one into integral of this that
is minus sin n x upon n square from zero to pi now this will be zero when x is pi and
when x is zero only this will be left so it is minus two upon pi n i am taking outside
it is only exists when x is pi so it is pi into minus one k to the power n ok pi pi cancel
out b n is nothing but b n is nothing but uh minus two by n into minus one k to the
power n ok now so what is a fourier series representation of this function what is the
function x minus x square so f x will be equal to a naught by two a naught is is this term
eta a naught by two is minus pi square upon three plus summation n from one to infinity
a n a n is minus four by n square minus one k to the power n cos n x plus summation n
from one to infinity b n b n is minus two k to the power two upon n minus one k to the
power n sin n x so what the series is this is minus pi square upon three minus four times
when you take n equal to one when n is one it is it is minus cos x when n is two it is
plus cos two x upon two square when n is three it is minus cos three x upon three square
and so on and it is minus two times when n is one it is minus sin x upon one plus when
n is two it is sin two x upon two when n is three it is minus sin three x upon three and
so on ok so now we want to formulate this series ok
this is the fourier representation of fourier representation of this function ok the first
part is over now we want to deduce this equal two pi square by twelve now it is one by one
square one by two square with alternate signs ok one by three square one by four square
with alternate sign and we dont want any term from this series so this we can obtain putting
x equal to zero when you put x equal to zero all these terms are cos x cos two x cos three
x all will be one and all these will be zero so put x equal to zero so if we put x equal
to zero the left hand side is zero it is minus pi square by three take minus common from
here it is four times it is one by one square minus one by two square plus one by three
square and so on and this is zero so what will be the series of this it is nothing but
one upon one square plus one by two square plus one by three square and so on is equal
to pi square by three will go here divide by four so it is pi square by twelve so hence
this result can be obtained ok so in the next class we will see that what how we can say
that what is the convergence theorem for the fourier series that we will see ok so
thank you very much